Problem: Divide the following complex numbers. $ \dfrac{10-20i}{4-3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4+3i}$ $ \dfrac{10-20i}{4-3i} = \dfrac{10-20i}{4-3i} \cdot \dfrac{{4+3i}}{{4+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(10-20i) \cdot (4+3i)} {(4-3i) \cdot (4+3i)} = \dfrac{(10-20i) \cdot (4+3i)} {4^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(10-20i) \cdot (4+3i)} {(4)^2 - (-3i)^2} = $ $ \dfrac{(10-20i) \cdot (4+3i)} {16 + 9} = $ $ \dfrac{(10-20i) \cdot (4+3i)} {25} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({10-20i}) \cdot ({4+3i})} {25} = $ $ \dfrac{{10} \cdot {4} + {-20} \cdot {4 i} + {10} \cdot {3 i} + {-20} \cdot {3 i^2}} {25} $ Evaluate each product of two numbers. $ \dfrac{40 - 80i + 30i - 60 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{40 - 80i + 30i + 60} {25} = \dfrac{100 - 50i} {25} = 4-2i $